Problem 1:
One of the most prominent spectral lines of hydrogen is the Hα
line, a bright red line with a wavelength of 656.1 nm.
What is the wavelength of the Hα line emitted from a star receding
from the observer with a speed of 3000 km/s?
Solution:
- Concepts:
Doppler shift
- Reasoning:
The star is receding, the spectral line is red-shifted.
- Details of the calculation:
f' = f[(1 - v/c)/(1 + v/c)]½ is the Doppler shift formula for
a receding source.
λ' = c/f' = (c/f)[(1 + v/c)/(1 - v/c)]½ = λ[(1 + v/c)/(1 -
v/c)]½.
λ' = (656.1 nm)[(1 + 10-2)/(1 - 10-2)]½ =
662.7 nm.
Problem 2:
Consider an electron-positron collider, a circular ring with a beam of electrons
and a beam of positrons traveling in opposite directions. The mass of the
electron is 0.5 MeV/c2.
(a) If you wanted to increase the speed of the particles in the beams
(accelerate them in the direction of their velocity) could you use electric
fields, magnetic fields, or either? Why?
(b) If each beam is accelerated to an energy of 100 GeV, and the ring has a
circumference of 1 km, how long would a trip around the ring take in the rest
frame of the electron?
Solution:
- Concepts:
The
proper time interval in an accelerating frame
- Reasoning:
The rest frame of the electron is not an inertial frame. We imagine an inertial
frame with identical instantaneous velocity v moving alongside the
electron for a time interval dτ measured in the imagined frame. The
corresponding time interval measured on the planet is dt = γdτ. Since
the acceleration of the electron is perpendicular to its velocity, its speed
with respect to the ground does not change, γ = (1 - v2/c2)-½ is constant.
- Details of the calculation:
(a) The Lorentz force, F = q(E + v × B).
The direction of the magnetic force acting on a charged particle is
perpendicular to the velocity of the particle. The resulting centripetal
acceleration changes the direction of the velocity, but not the speed. An
electric field component in the direction of the velocity changes the speed of
the particle.
(b) Let T be the time it takes the electron to complete one revolution in the lab
frame.
E = 100 GeV = γmc2 = γ(0.5*10-3) GeV, γ = 2*105.
v ≈ c, T = 2πr/c = (1000 m)/(3*108 m/s) = 3.33*10-6 s.
In the rest frame of the electron, T' = T/γ = 1.67*10-11 s.
Problem 3:
A neutral pi meson with energy twice its rest mass (E = 2mc2) is moving along
the x-axis with respect to observer O. It decays into two photons.
In the meson's rest frame the photons are emitted in opposite directions along
the y' axis. Find the energy of the photons in both the pi-meson's rest frame
and in the observer's frame, and the emission angles of the photons in the
observer's frame.
Solution:
- Concepts:
Relativistic energy and momentum conservation
- Reasoning:
The decay of a particle is a relativistic problem. In relativistic
"collisions" energy and momentum are always conserved.
- Details of the calculation:
In the pi-meson's rest frame the energy of each photon is hf = ½mc2,
one photon has momentum hf/c j and the other has momentum -hf/c j.
Since γ = 2, the pi meson moves with speed v = (√3)c/2 in the observer's
frame.
Since the photons are emitted in the pi meson's rest frame perpendicular to
the direction of the relative velocity of the frames, their energies hf' in the observers frame
are equal. Their momentum vectors lie in the xy-plane and make angles
+φ and -φ with the x-axis, respectively.
In the observer's frame we have from energy and momentum conservation
γmc2 =
2hf', hf' = mc2, hf'cosφ/c = ½γmv = m(√3)c/2.
cosφ =
(√3)/2. φ = 30o.
The angle between the emission directions of the photons is 60o.
Problem 4:
Two spaceships move in opposite directions in a circle of radius R with
constant angular velocities ω1 and ω2 as seen by
an observer in an inertial frame. When the ships
meet for the first time, their captains synchronize their clocks. When they meet again, whose clock
will be delayed and by how much?
Solution:
- Concepts:
Relativistic kinematics, the proper time interval
- Reasoning:
The ships are not inertial frames. But they move with constant speed with
respect to an inertial frame (frame 1) with its origin at the center of the
circle. When a clock in the accelerating frame measures a proper time
interval between two events, the clock in the inertial frame will measure a
longer time interval between the two events.
- Details of the calculation:
Each ship frame is not an inertial frame. For a ship moving with speed v,
we imagine an inertial frame with identical instantaneous velocity v moving alongside the
planet for a time interval dτ measured in the imagined frame. The
corresponding time interval measured in frame 1 is dt = γdτ. Since
the acceleration of the planet is perpendicular to its velocity, its speed
with respect to the planet does not change, γ = (1 - v2/c2)-½ is constant.
For the observer in the inertial frame a time interval Δt elapses before
the ships meet again.
ω1Δt + ω2Δt = 2π, Δt = 2π/(ω1
+ ω2).
Ship 1: v1 = ω1R,
γ1 = (1 - ω12R2/c2)-½,
τ1 = Δt/γ1
= Δt(1 - ω12R2/c2)½.
Ship 2: v2 = ω2R, γ2
= (1 - ω22R2/c2)-½,
τ2 = Δt/γ2
= Δt(1 - ω22R2/c2)½.
τ1 - τ2 = [2π/(ω1 + ω2)][(1
- ω12R2/c2)½ - (1
- ω22R2/c2)½].
If ω1 < ω2 , then τ1 - τ2
is positive. More time has elapsed on ship 1 than on ship 2.
Problem 5:
A spaceship is approaching its home planet with velocity v = 0.8c i, when
radio waves arriving from two stations on the planet signal that the stations
are under attack at the same time (t = 0) recorded on the planet. The
captain of the ship knows that station 1 has planet coordinate x = 0, and station
2 has planet coordinate x = L = 104 km.
What is the time difference between the two attacks in the ship's reference
frame, and which station was attacked first.
Solution: